SATHEE: Differential Equations 2 Question 3

Differential Equations 2 Question 3

3. Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, x \in - \frac{π}{2}, \frac{π}{2}$ , such that $y (0) = 1$ . Then

(2019 Main, 10 April II)

(a) $y^{'} \frac{π}{4} - y^{'} - \frac{π}{4} = π - \sqrt{2}$ (b)

$y^{'} \frac{π}{4} + y^{'} - \frac{π}{4} = - \sqrt{2}$

(d) $y \frac{π}{4} - y - \frac{π}{4} = \sqrt{2}$

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Answer:

Correct Answer: 3. (a)

Solution:

Given differential equation is

$\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x$ , which is linear differential equation in the form of $\frac{d y}{d x} + P y = Q$ .

Here, $P = \tan x$ and $Q = 2 x + x^{2} \tan x$

$∴ I F = e^{\int \tan x d x} = e^{\log_{e} (\sec x)} = \sec x$

Now, solution of linear differential equation is given as

$\begin{aligned} y \times I F = \int (Q \times I F) d x + C \\ ∴ y (\sec x) = \int (2 x + x^{2} \tan x) \sec x d x + C \\ = \int (2 x \sec x) d x + \int x^{2} \sec x \tan x d x + C \\ ∵ \int x^{2} \sec x \tan x d x = x^{2} \sec x - \int (2 x \sec x) d x \end{aligned}$

Therefore, solution is

$y \sec x = 2 \int x \sec x d x + x^{2} \sec x - 2 \int x \sec x d x + C$

$\Rightarrow y \sec x = x^{2} \sec x + C \dots$ (i)

$∵ y (0) = 1 \Rightarrow 1 (1) = 0 (1) + C \Rightarrow C = 1$

Now, $y = x^{2} + \cos x$

[from Eq. (i)]

and $y^{'} = 2 x - \sin x$

According to options,

$y^{'} \frac{π}{4} - y^{'} \frac{- π}{4} = 2 \frac{π}{4} - \frac{1}{\sqrt{2}}$

$- 2 - \frac{π}{4} + \frac{1}{\sqrt{2}} = π - \sqrt{2}$

and $y^{'} \frac{π}{4} + y^{'} - \frac{π}{4} = 2 \frac{π}{4} - \frac{1}{\sqrt{2}} + 2 - \frac{π}{4} + \frac{1}{\sqrt{2}} = 0$

and $y \frac{π}{4} + y - \frac{π}{4} = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} + \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} = \frac{π^{2}}{4} + \sqrt{2}$

and $y \frac{π}{4} - y - \frac{π}{4} = \frac{π^{2}}{16} + \frac{1}{\sqrt{2}} - \frac{π^{2}}{16} - \frac{1}{\sqrt{2}} = 0$

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Differential Equations 2 Question 3

3. Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, x \in - \frac{π}{2}, \frac{π}{2}$ , such that $y (0) = 1$ . Then

Answer:

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Differential Equations 2 Question 3

3. Let y=y(x) be the solution of the differential equation, dydx+ytan⁡x=2x+x2tan⁡x,x∈−π2,π2, such that y(0)=1. Then

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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3. Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, x \in - \frac{π}{2}, \frac{π}{2}$ , such that $y (0) = 1$ . Then